What will happen when you perform an operation between an int and a float in a mixed expression?

In a mixed expression where an operation involves both an integer and a floating-point number, implicit type conversion happens to ensure that the operation can be performed accurately. In this case, the integer will be converted to a floating-point value. This conversion allows the arithmetic operation to take place without losing any decimal precision that might be associated with the float.

When an integer is converted to a float, its value remains the same, but it is represented in a way that can accommodate decimal points. This is crucial because if the integer were to be ignored or converted to an int during the operation, it could lead to loss of precision, especially in cases where results may yield a decimal value.

The conversion is a part of the rules of type promotion in C++, where the data type that can hold a larger range or more precision (in this case, float) takes precedence over the other. Thus, by converting the integer to a float, both values can be treated consistently in the operation, resulting in a float output.